Complete a Write-up
on your Web Page for the following investigation. This should
be individual work.

A. Consider any triangle
ABC. Select a point P inside the triangle and draw lines AP, BP,
and CP extended to their intersections with the opposite sides
in points D, E, and F respectively.

Explore (AF)(BD)(EC)
and (FB)(DC)(EA) for various triangles and various locations of
P.

(AF)(BD)(EC) = (FB)(DC)(EA),
regardless of where the point P is placed. My hypothesis is that
this is true because of some ratio of the corresponding sides
of the triangle ABC.

Using the various
line segments through the point P, there are 6 different triangles
to look at. We can assume that several are similar which supports
that a common ratio of the sides, may exist. Using parallel lines,
we can find the following two similar triangles:

Based on the
parallel lines shown, the following pairs of angles are equal.
<ZEC = <PEA and <ZCE = <EAP. Therefore, the two shaded
triangles must be similar which implies that AP/CZ = AE/EC. Again,
using parallel lines, we can show that the following two shaded
triangles are parallel, as well. <AFP = <XFB, and <FBX
= <FAP, such that the following proportion is true, AF/BF =
AP/XB.

Because of corresponding
angles and the common angle, the following two shaded triangles
are similar:

and the following
two triangles are similar for the same reason.

Such that we
get the following proportion based on Ceva's Theorem:

AF/BF * BC/CD
* CD/BC * CE/AE =

AP/BX * DP/CY
* BX/DP * CY/AP.

We can simplify
this and get the desired result, (AF)(BD)(EC)/(FB)(DC)(EA) = 1.

The ratio of the areas
of the two triangles will always be greater than or equal to 4.
To see this click
here.